Question

Suppose that X is a normal random variable and E(X) = −3. Find Var(X) if P(−7 < X < 1) = 0.7888.

Answer #1

For normal distribution, P(X < A) = P(Z < (A - )/)

E(X), = -3

Var(X), ^{2}
= ?

P(-7 < X < 1) = 0.7888

-3 is the midpoint of the interval (-7, 1) and it is the mean of the distribution.

Therefore P(-7 < X < -3) = P(-3 < X < 1) = 0.7888/2 = 0.3944

P(-3 < X < 1) = 0.3944

P(X < 1) - P(X < -3) = 0.3944

P(Z < (1 - -3)/) + 0.5 = 0.3994

P(Z < 4/) = 0.8994

Take the z score corresponding to 0.8994 from standard normal distribution table

4/ = 1.28

= 3.125

^{2}
= 3.13^{2} = 9.8

Var(X) = **9.8**

Let X be a binomial random variable with E(X) = 7 and Var(X) =
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(a) [5 pts] Find the parameters n and p for the binomial
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n =
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(i) Find approximately (using the Z-table) what is Var(X).
(ii) Find the value c such that P(X > c) = 0.1.

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p (1) = P (X = 1) = 0.2
p (2) = P (X = 2) = 0.1
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a. Find E(X^2) .
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a. E(X)
b. Var(X)
c. P(1<X<3)
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(b) For T = 2X + Y find µT , σT
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p(x>1)=
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Value x of X P(X=x)
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6 0.55
7 0.10
Find the expectation E (X) and variance Var (X) of X. (If
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